Problem: The angle $\theta_1$ is located in Quadrant $\text{IV}$, and $\sin(\theta_1)=-\dfrac{10}{13}$. What is the value of $\cos(\theta_1)$ ? Express your answer exactly. $\cos(\theta_1)=$
The Strategy We can use the Pythagorean identity, $\cos^2(\theta)+\sin^2(\theta)=1$, to solve for $\cos(\theta_1)$ from $\sin(\theta_1)$ or vice versa. [How did we get the Pythagorean identity?] In this case, we can find $\cos(\theta_1)$ by doing the following. Find $\cos^2(\theta_1)$ using $\sin(\theta_1)$ and the Pythagorean identity. [What does this notation mean?] Determine $\cos(\theta_1)$ by considering the quadrant of $\theta_1$. Finding $\cos^2(\theta_1)$ Let's plug in $\sin(\theta_1)=-\dfrac{10}{13}$ into the equation $\cos^2(\theta)+\sin^2(\theta)=1$ to solve for $\cos^2(\theta_1)$. $\begin{aligned}\cos^2(\theta_1)+\sin^2(\theta_1)&=1 \\\\\cos^2(\theta_1)&={1-\sin^2(\theta_1)} \\&=1-\left(-\dfrac{10}{13}\right)^2 \\&=\dfrac{69}{169}\end{aligned}$ Finding $\cos(\theta_1)$ Since $\theta_1$ is in Quadrant $\text{IV}$, $\cos\theta_1$ is positive. $\begin{aligned}\cos(\theta_1)&=\sqrt{\cos^2(\theta_1)} \\&=\sqrt{\dfrac{69}{169}} \\&=\dfrac{\sqrt{69}}{13}\end{aligned}$ Summary $\cos(\theta_1)=\dfrac{\sqrt{69}}{13}$